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Three-dice-are-rolled-The-number-of-possible-outcomes-in-which-at-least-one-die-shows-5-is-




Question Number 25334 by arvindbrar26068@gmail.com last updated on 08/Dec/17
Three dice are rolled. The number of  possible outcomes in which at least  one die shows 5 is
$$\mathrm{Three}\:\mathrm{dice}\:\mathrm{are}\:\mathrm{rolled}.\:\mathrm{The}\:\mathrm{number}\:\mathrm{of} \\ $$$$\mathrm{possible}\:\mathrm{outcomes}\:\mathrm{in}\:\mathrm{which}\:\mathrm{at}\:\mathrm{least} \\ $$$$\mathrm{one}\:\mathrm{die}\:\mathrm{shows}\:\mathrm{5}\:\mathrm{is} \\ $$
Answered by ajfour last updated on 08/Dec/17
coeff. of x^1 +coeff. of x^2 +  coeff. of x^3  in (5+x)^3     =75+15+1 =91 .  or  =216−125 =91 .
$${coeff}.\:{of}\:{x}^{\mathrm{1}} +{coeff}.\:{of}\:{x}^{\mathrm{2}} + \\ $$$${coeff}.\:{of}\:{x}^{\mathrm{3}} \:{in}\:\left(\mathrm{5}+{x}\right)^{\mathrm{3}} \\ $$$$\:\:=\mathrm{75}+\mathrm{15}+\mathrm{1}\:=\mathrm{91}\:. \\ $$$${or} \\ $$$$=\mathrm{216}−\mathrm{125}\:=\mathrm{91}\:. \\ $$

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